力扣算法题—079单词搜索【DFS】

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.

 #include "_000库函数.h"

 //此问题是一个DFS问题，每个单词可以像上下左右四个方向进行下一步行走

 class Solution {
 public:
     bool exist(vector<vector<char>>& board, string word) {
         if (board.empty() || board[].empty())return false;
         int m = board.size(), n = board[].size();
         vector<vector<bool>>visted(m, vector<bool>(n, false));//记录是否被访问过
         for (int i = ; i < m; ++i) {
             for (int j = ; j < n; ++j) {
                 if (search(board, visted, , i, j, word))return true;//每个单词单元都进行上下左右进行搜索！
             }
         }
     }

     bool search(vector<vector<char>>& board, vector<vector<bool>>& visted, int idx, int i, int j, string word) {
         if (idx == word.size())return true;
         int m = board.size(), n = board[].size();
         if (i <  || j <  || i >= m || j >= n || visted[i][j] || board[i][j] != word[idx])return false;
         visted[i][j] = true;
         bool res = (search(board, visted, idx + , i - , j, word) ||
                     search(board, visted, idx + , i + , j, word) ||
                     search(board, visted, idx + , i, j - , word) ||
                     search(board, visted, idx + , i, j + , word));
         visted[i][j] = false;//回溯
         return res;
     }
 };

 void T079() {
     Solution s;
     vector<vector<char>> board;
     string word;
     board = { {'A', 'B', 'C', 'E'}, {'S', 'F', 'C', 'S'}, {'A', 'D', 'E', 'E'} };
     word = "ABCCED";
     cout << s.exist(board, word) << endl;

 }